About the project

This is a course about statistics with open source tools: R, RStudio, and so on.

This course is massive!

Every week there will be a submission with a deadline on the following Sunday at 23:55. Late submissions will not be accepted.

After that, three other projects will be assigned for peer review. The deadline for the peer reviews is Wednesday the same week at 23:55.

The exercises and the peer reviews are all that is required for this course.

Here is the link to my GitHub repository.

Regression and model validation

Load the wrangled data and take a look at it.

learning2014 <- read.table("data/learning2014.tsv", sep = "\t")
str(learning2014)
## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
##  $ age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ attitude: num  3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ points  : int  25 12 24 10 22 21 21 31 24 26 ...

Plot the data.

library(ggplot2)

# initialize plot with data and aesthetic mapping
p1 <- ggplot(learning2014, aes(x = attitude, y = points, col = gender))

# define the visualization type (points)
p2 <- p1 + geom_point()

# add a regression line
p3 <- p2 + geom_smooth(method = "lm")

# add a main title and draw the plot
p4 <- p3 + ggtitle("Student's attitude versus exam points")

# draw the plot
p4

There is a positive correlation between attitude and points. There is no obvious difference between the two genders. Interestingly, there are only two genders.

Now, plot all possible pairs of variables as scatter plots:

pairs(learning2014[-1])

This is not very informative, since the plots are very small and there is no regression line to help us imagine which way the correlation goes.

Drawing more advanced plots:

library(GGally)
library(ggplot2)

# create a more advanced plot matrix with ggpairs()
p <- ggpairs(learning2014, mapping = aes(col = gender, alpha = .3), lower = list(combo = wrap("facethist", bins = 20)))

# draw the plot
p

Age seems to follow a Poisson distribution, probably because the subjects are students. The rest of the variables seem to have a normal distribution, as expected.

Apparently, attitude and points have the highest correlation by far.

In addition, there is a negative correlation between surf and deep.

There is a slight negative correlation between surf and points, surf and age, and surf and stra, as well as a positive correlation between stra and points.

Now, a regression analysis using the three variables that had the highest individual correlation with points:

# create a regression model with three explanatory variables
my_model2 <- lm(points ~ attitude + stra + surf, data = learning2014)

# print out a summary of the model
summary(my_model2)
## 
## Call:
## lm(formula = points ~ attitude + stra + surf, data = learning2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.1550  -3.4346   0.5156   3.6401  10.8952 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.0171     3.6837   2.991  0.00322 ** 
## attitude      3.3952     0.5741   5.913 1.93e-08 ***
## stra          0.8531     0.5416   1.575  0.11716    
## surf         -0.5861     0.8014  -0.731  0.46563    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared:  0.2074, Adjusted R-squared:  0.1927 
## F-statistic: 14.13 on 3 and 162 DF,  p-value: 3.156e-08

There seems to be a statistically significant relationship between the chosen variables. Here are the diagnostic plots:

par(mfrow = c(2,2))
plot(my_model2, which = c(1, 2, 5))

Nothing looks out of the ordinary in these diagnostic plots. There is no reason not to trust the hypothesis that all three chosen variables – attitude, stra, and surf, are explanatory variables for points. The correlation of course is not incredibly strong.

Logistic regression

Data

The data was joined by using following columns as surrogate identifiers for students: school, sex, age, address, famsize, Pstatus, Medu, Fedu, Mjob, Fjob, reason, nursery, internet.

Defined two new variables: alc_use and high_use.

alc <- read.csv("data/alc.csv")
str(alc)
## 'data.frame':    382 obs. of  35 variables:
##  $ school    : Factor w/ 2 levels "GP","MS": 1 1 1 1 1 1 1 1 1 1 ...
##  $ sex       : Factor w/ 2 levels "F","M": 1 1 1 1 1 2 2 1 2 2 ...
##  $ age       : int  18 17 15 15 16 16 16 17 15 15 ...
##  $ address   : Factor w/ 2 levels "R","U": 2 2 2 2 2 2 2 2 2 2 ...
##  $ famsize   : Factor w/ 2 levels "GT3","LE3": 1 1 2 1 1 2 2 1 2 1 ...
##  $ Pstatus   : Factor w/ 2 levels "A","T": 1 2 2 2 2 2 2 1 1 2 ...
##  $ Medu      : int  4 1 1 4 3 4 2 4 3 3 ...
##  $ Fedu      : int  4 1 1 2 3 3 2 4 2 4 ...
##  $ Mjob      : Factor w/ 5 levels "at_home","health",..: 1 1 1 2 3 4 3 3 4 3 ...
##  $ Fjob      : Factor w/ 5 levels "at_home","health",..: 5 3 3 4 3 3 3 5 3 3 ...
##  $ reason    : Factor w/ 4 levels "course","home",..: 1 1 3 2 2 4 2 2 2 2 ...
##  $ nursery   : Factor w/ 2 levels "no","yes": 2 1 2 2 2 2 2 2 2 2 ...
##  $ internet  : Factor w/ 2 levels "no","yes": 1 2 2 2 1 2 2 1 2 2 ...
##  $ guardian  : Factor w/ 3 levels "father","mother",..: 2 1 2 2 1 2 2 2 2 2 ...
##  $ traveltime: int  2 1 1 1 1 1 1 2 1 1 ...
##  $ studytime : int  2 2 2 3 2 2 2 2 2 2 ...
##  $ failures  : int  0 0 2 0 0 0 0 0 0 0 ...
##  $ schoolsup : Factor w/ 2 levels "no","yes": 2 1 2 1 1 1 1 2 1 1 ...
##  $ famsup    : Factor w/ 2 levels "no","yes": 1 2 1 2 2 2 1 2 2 2 ...
##  $ paid      : Factor w/ 2 levels "no","yes": 1 1 2 2 2 2 1 1 2 2 ...
##  $ activities: Factor w/ 2 levels "no","yes": 1 1 1 2 1 2 1 1 1 2 ...
##  $ higher    : Factor w/ 2 levels "no","yes": 2 2 2 2 2 2 2 2 2 2 ...
##  $ romantic  : Factor w/ 2 levels "no","yes": 1 1 1 2 1 1 1 1 1 1 ...
##  $ famrel    : int  4 5 4 3 4 5 4 4 4 5 ...
##  $ freetime  : int  3 3 3 2 3 4 4 1 2 5 ...
##  $ goout     : int  4 3 2 2 2 2 4 4 2 1 ...
##  $ Dalc      : int  1 1 2 1 1 1 1 1 1 1 ...
##  $ Walc      : int  1 1 3 1 2 2 1 1 1 1 ...
##  $ health    : int  3 3 3 5 5 5 3 1 1 5 ...
##  $ absences  : int  5 3 8 1 2 8 0 4 0 0 ...
##  $ G1        : int  2 7 10 14 8 14 12 8 16 13 ...
##  $ G2        : int  8 8 10 14 12 14 12 9 17 14 ...
##  $ G3        : int  8 8 11 14 12 14 12 10 18 14 ...
##  $ alc_use   : num  1 1 2.5 1 1.5 1.5 1 1 1 1 ...
##  $ high_use  : logi  FALSE FALSE TRUE FALSE FALSE FALSE ...

Predictors of high alcohol consumption

The following four variables were chosen:

goout, sex, studytime, and romantic.

The hypothesis is that going out leads to dringking and males drink more because they are heavier on average. Furthermore, drinking and going out leaves less time for studying. Being in a relationship should reduce alcohol consumption since there is no need to get wasted and meet people.

Data exploration

Explore the variables of interest:

library(tidyr)
library(dplyr)
library(ggplot2)

alc %>% group_by(high_use) %>% summarise(count = n(), mean_goout=mean(goout),
                                         mean_studytime=mean(studytime))
## # A tibble: 2 x 4
##   high_use count mean_goout mean_studytime
##   <lgl>    <int>      <dbl>          <dbl>
## 1 FALSE      268       2.85           2.15
## 2 TRUE       114       3.72           1.77
alc %>% group_by(high_use, sex) %>% summarise(count = n())
## # A tibble: 4 x 3
## # Groups:   high_use [?]
##   high_use sex   count
##   <lgl>    <fct> <int>
## 1 FALSE    F       156
## 2 FALSE    M       112
## 3 TRUE     F        42
## 4 TRUE     M        72
alc %>% group_by(high_use, romantic) %>% summarise(count = n())
## # A tibble: 4 x 3
## # Groups:   high_use [?]
##   high_use romantic count
##   <lgl>    <fct>    <int>
## 1 FALSE    no         180
## 2 FALSE    yes         88
## 3 TRUE     no          81
## 4 TRUE     yes         33
g_goout <- ggplot(alc, aes(x = goout, fill=high_use)) +
  geom_bar() + xlab("Going out with friends") +
  ggtitle("Going out with friends from 1 (very low) to 5 (very high) by alcohol use")

g_studytime <- ggplot(alc, aes(x = studytime, fill=high_use)) +
  geom_bar() + xlab("Weekly study time") +
  ggtitle("Weekly study time [1 (<2 hours), 2 (2 to 5 hours), 3 (5 to 10 hours), or 4 (>10 hours)] by alchol use")

g_sex <- ggplot(alc, aes(x = sex, fill=high_use)) +
  geom_bar() +
  ggtitle("Sex by alcohol use")

g_romantic <- ggplot(alc, aes(x = romantic, fill=high_use)) +
  geom_bar() +
  ggtitle("With a romantic relationship (yes/no) by alcohol use")

# Arrange the plots into a grid
library("gridExtra")
grid.arrange(g_goout, g_studytime, g_sex, g_romantic, ncol=2, nrow=2)

In summary, all parts of the hypothesis seem to be correct.

Fitting a logistic regression model

Fit a logistic regression model using high_use as the target variable and goout, studytime, sex, and romantic as explanatory variables.

m <- glm(high_use ~ goout + studytime + sex + romantic, data = alc, family = "binomial")

Variables goout, studytime and sex are associated with alcohol consumption.

High alcohol consumption is associated with going out (as expected).

Males who drink less study more.

Summary of the model:

summary(m)
## 
## Call:
## glm(formula = high_use ~ goout + studytime + sex + romantic, 
##     family = "binomial", data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.7365  -0.8114  -0.5009   0.9081   2.6642  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -2.6988     0.5712  -4.725 2.30e-06 ***
## goout         0.7536     0.1187   6.350 2.15e-10 ***
## studytime    -0.4774     0.1683  -2.837  0.00456 ** 
## sexM          0.6657     0.2585   2.576  0.01000 *  
## romanticyes  -0.1424     0.2699  -0.528  0.59767    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 393.67  on 377  degrees of freedom
## AIC: 403.67
## 
## Number of Fisher Scoring iterations: 4

Coefficients of the model as odds ratios and their confidence intervals:

or <- coef(m) %>% exp
ci <- confint(m) %>% exp
cbind(or, ci)
##                     or      2.5 %    97.5 %
## (Intercept) 0.06728696 0.02129867 0.2010636
## goout       2.12456419 1.69404697 2.7003422
## studytime   0.62040325 0.44145946 0.8558631
## sexM        1.94589655 1.17538595 3.2443631
## romanticyes 0.86724548 0.50714091 1.4648961

In summary, 1 unit increase in goout is associated with 2.1 increase in likelihood of high alcohol consumption.

1 unit increase in studytime is associated with 0.6 lower likelihood of high alcohol consumption.

Being male is associated with 1.9 times increase in likelihood of high alcohol consumption compared to being female.

Being in a romantic relationship is not significantly associated with a change in likelihood of high alcohol consumption, so the hypothesis was wrong.

Performance of the model

Fit a logistic model with the explanatory variables that were statistically significantly associated to high or low alcohol consumption:

m <- glm(high_use ~ goout + studytime + sex, data = alc, family = "binomial")

Prediction performance of the model:

probability <- predict(m, type="response")
alc <- mutate(alc, probability=probability)
alc <- mutate(alc, prediction=probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction)
##         prediction
## high_use FALSE TRUE
##    FALSE   250   18
##    TRUE     76   38

The model is better at predicting low alcohol consumption than high alcohol consumption.

Visualizing the class, the predicted probabilities, and the predicted class:

g <- ggplot(alc, aes(x = probability, y = high_use, col=prediction))
g + geom_point()

Calculate the total proportion of misclassified individuals using the regression model. Use a simple guessing strategy where everyone is classified to be in the most prevalent class:

loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}

loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2460733
loss_func(class = alc$high_use, prob = 0)
## [1] 0.2984293

Using the regression model, 24.6% of the individuals are misclassified, compared to 29.8 % of misclassified individuals if guessing that everybody belongs to the low use of alcohol class. The model seems to provide modest improvement to the simple guess of the most prevalent class.

Cross-validation

Perform 10-fold cross-validation of the model to estimate the performance of the model on unseen data. The performance of the model is measured with proportion of misclassified individuals. The mean prediction error in the test set:

library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
cv$delta[1]
## [1] 0.2460733

The mean prediction error in the test set is 0.25, marginally better than the performance of the model in the DataCamp exercises, with a mean prediction error of 0.26 in the test set.

Models with different number of predictors

Construct models with different number of predictors and calculate the test set and training set prediction errors:

predictors <- c('school', 'sex', 'age', 'address', 'famsize', 'Pstatus', 'Medu', 'Fedu', 'Mjob', 'Fjob', 'reason', 'nursery', 'internet', 'guardian', 'traveltime', 'studytime', 'failures', 'schoolsup', 'famsup', 'paid', 'activities', 'higher', 'romantic', 'famrel', 'freetime', 'goout', 'health', 'absences', 'G1', 'G2', 'G3')

# Fit several models and record the test and traingin errors
# 1) Use all of the predictors.
# 2) Drop one predictor and fit a new model.
# 3) Continue until only one predictor is left in the model.

test_error <- numeric(length(predictors))
training_error <- numeric(length(predictors))

for(i in length(predictors):1) {
  model_formula <- paste0("high_use ~ ", paste(predictors[1:i], collapse = " + "))
  glmfit <- glm(model_formula, data = alc, family = "binomial")
  cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
  test_error[i] <- cv$delta[1]
  training_error[i] <- loss_func(alc$high_use, predict(glmfit,type="response"))
}

data_error <- rbind(data.frame(n_predictors=1:length(predictors),
                               prediction_error=test_error,
                               type = "test error"),
                    data.frame(n_predictors=1:length(predictors),
                               prediction_error=training_error,
                               type = "training error"))


g <- ggplot(data_error, aes(x = n_predictors, y = prediction_error, col=type))
g + geom_point()

Clustering and classification

Load Boston dataset from the MASS package:

library(corrplot)
library(dplyr)
library(MASS)
data("Boston")
str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

The dataset has 14 variables and 506 observations. Full details can be found in the dataset’s documentation.

Data exploration

Summary of the variables in the dataset:

summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

Plot the variables and explore the data:

library(GGally)
library(ggplot2)
p <- ggpairs(Boston, mapping = aes(alpha=0.3), 
             lower = list(combo = wrap("facethist", bins = 20)))
p

Correlation of the variables:

cor(Boston) %>% corrplot(method = "circle", type = "upper", cl.pos = "b", tl.pos = "d")

Data wrangling

Scaling the dataset so that the average is \(0\) and standard deviation is \(1\):

\[x_{scaled}=\frac{x - \mu_{x}}{\sigma_{x}}\] where \(\mu_{x}\) is the mean of \(x\) and \(\sigma_{x}\) is the standard deviation of \(x\).

boston_scaled <- scale(Boston) %>% as.data.frame()
summary(boston_scaled)
##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865

Create a factor variable crime from the crim by dividing crim by quartiles to “low”, “med_low”, “med_high” and “high” categories:

bins <- quantile(boston_scaled$crim)
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, 
             label=c("low", "med_low", "med_high", "high"))
boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled <- data.frame(boston_scaled, crime)

Divide the dataset to training and test sets so that 80% belongs to the training set and 20% belongs to the test set:

set.seed(1)
train.idx <- sample(nrow(boston_scaled), size = 0.8 * nrow(boston_scaled))
train <- boston_scaled[train.idx,]
test <- boston_scaled[-train.idx,]

Linear discriminant analysis

Fit the linear discriminant analysis (LDA) on the training set using the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables:

lda.fit <- lda(crime ~ ., data = train)

The LDA bi-plot:

lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)) {
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]],
         col = color, length = arrow_heads)
  text(myscale * heads[,choices],
       labels = row.names(heads), 
       cex = tex, col = color, pos = 3)
}
classes <- as.numeric(train$crime)
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 2)

Use the fitted LDA model to predict the categorical crime rate in the test set. Cross-tabulate the observed classes and the predicted classes:

correct_classes <- test$crime
test <- dplyr::select(test, -crime)
lda.pred <- predict(lda.fit, newdata = test)
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       15      11        1    0
##   med_low    8      20        2    0
##   med_high   1       9       16    0
##   high       0       0        0   19

Model seems to perform perfectly at predicting the “high” class and predicts the other classes reasonably well. The prediction accuracy is worst for the “low” class. The model misclassifies a large proportion of the “low” observations as “med_low”.

K-means clustering

Reload the Boston dataset and standardize it as above. Calculate the Euclidean distance between the observations:

data("Boston")
boston_scaled <- scale(Boston) %>% as.data.frame()
dist_eu <- dist(boston_scaled)
summary(dist_eu)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

Run the k-means algorithm with 3 clusters and visualize the results:

# seeded above
km <-kmeans(boston_scaled, centers = 3)
pairs(boston_scaled, col = km$cluster)

Calculate the total of within cluster sum of squares (TWCSS) when the number of cluster changes from 1 to 10:

k_max <- 10
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})
qplot(x = 1:k_max, y = twcss, geom = 'line')

The optimal number of clusters is when the total WCSS drops radically. Based on the graph, 2 seems to be the optimal number.

Perform k-means with 2 clusters and visualize the results:

km <-kmeans(boston_scaled, centers = 2)
pairs(boston_scaled, col = km$cluster)

LDA of the k-means clusters

Perform k-means clustering with 3 clusters on the scaled Boston dataset. Use the cluster assignments as the target variable for LDA analysis:

km <-kmeans(boston_scaled, centers = 3)
boston_scaled$kmeans_cluster <- km$cluster
lda.fit <- lda(kmeans_cluster ~ ., data = boston_scaled)

The LDA bi-plot:

plot(lda.fit, dimen = 2, col = boston_scaled$kmeans_cluster, pch = boston_scaled$kmeans_cluster)
lda.arrows(lda.fit, myscale = 2)

Based on the biplot, the most influential linear separators are age, dis, rad, and tax.